Group Law Elliptic Curve with Divisor Class Group
$begingroup$
Let $E$ be an elliptic curve and $k$ a field. It is well know that $E(k)$ has an (additive) group structure and indeed there are a lot of sources describing what geometrically there is going on.
The point of my interest is to find a derivation of this group law in language of divisors - especially using properties of the divisor class group.
Indeed the divisors $Div(E)$ are formal sums $sum_{P in E(k)} n_P (P)$ with $n_P in mathbb{Z}$ and the principal divisors $div(f) = sum_P ord_P(f) (P)$ form a subgroup of $Div(E)$; denote it by $PrDiv(E)$.
The divisor class group is the quotient $Cl(E)= Div(E)/PrDiv(E)$.
Obviosly we can define canonically a map $E(k) to Cl(E), p to (P)-(O)$ where $O$ is the special point (=neutral element).
How to show that this map determine the group law on $E(k)$.
algebraic-geometry elliptic-curves
asked Jan 11 at 21:24
KarlPeterKarlPeter
6081315
$endgroup$
|
show 7 more comments
$begingroup$
Let $E$ be an elliptic curve and $k$ a field. It is well know that $E(k)$ has an (additive) group structure and indeed there are a lot of sources describing what geometrically there is going on.
The point of my interest is to find a derivation of this group law in language of divisors - especially using properties of the divisor class group.
Indeed the divisors $Div(E)$ are formal sums $sum_{P in E(k)} n_P (P)$ with $n_P in mathbb{Z}$ and the principal divisors $div(f) = sum_P ord_P(f) (P)$ form a subgroup of $Div(E)$; denote it by $PrDiv(E)$.
The divisor class group is the quotient $Cl(E)= Div(E)/PrDiv(E)$.
Obviosly we can define canonically a map $E(k) to Cl(E), p to (P)-(O)$ where $O$ is the special point (=neutral element).
How to show that this map determine the group law on $E(k)$.
algebraic-geometry elliptic-curves
asked Jan 11 at 21:24
KarlPeterKarlPeter
6081315
$endgroup$
2
$begingroup$
$(O)$ isn't a principal divisor. Principal divisors have degree zero.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 21:27
$begingroup$
@LordSharktheUnknown: ah sure sure, $(O)-(O)$ should be the prinipal one and is obviously one. Thank you. But the point is why in $Cl(E)$ the group law is already determined.
$endgroup$
– KarlPeter
Jan 11 at 21:31
$begingroup$
This is Proposition III.3.4 (p. 61) in Silverman. Have you looked there?
$endgroup$
– André 3000
Jan 11 at 21:33
$begingroup$
Every degree zero divisor is congruent, modulo principal divisors, to exactly one of the form $(P)-(O)$.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 21:33
$begingroup$
It suffices to show (or define) that $P+Q-R-O$ is a principal divisor iff $P + Q= R$ in the group law. From this then $sum_j P_j-Q_j = R-O+div(f)$ where $R = sum_j P_j-Q_j$ in the group law and $div(f) = O-R+sum_j P_j-Q_j$ where $f$ is meromorphic
$endgroup$
– reuns
Jan 11 at 22:05
|
show 7 more comments
$begingroup$
Let $E$ be an elliptic curve and $k$ a field. It is well know that $E(k)$ has an (additive) group structure and indeed there are a lot of sources describing what geometrically there is going on.
The point of my interest is to find a derivation of this group law in language of divisors - especially using properties of the divisor class group.
Indeed the divisors $Div(E)$ are formal sums $sum_{P in E(k)} n_P (P)$ with $n_P in mathbb{Z}$ and the principal divisors $div(f) = sum_P ord_P(f) (P)$ form a subgroup of $Div(E)$; denote it by $PrDiv(E)$.
The divisor class group is the quotient $Cl(E)= Div(E)/PrDiv(E)$.
Obviosly we can define canonically a map $E(k) to Cl(E), p to (P)-(O)$ where $O$ is the special point (=neutral element).
How to show that this map determine the group law on $E(k)$.
algebraic-geometry elliptic-curves
asked Jan 11 at 21:24
KarlPeterKarlPeter
6081315
$endgroup$
Let $E$ be an elliptic curve and $k$ a field. It is well know that $E(k)$ has an (additive) group structure and indeed there are a lot of sources describing what geometrically there is going on.
The point of my interest is to find a derivation of this group law in language of divisors - especially using properties of the divisor class group.
Indeed the divisors $Div(E)$ are formal sums $sum_{P in E(k)} n_P (P)$ with $n_P in mathbb{Z}$ and the principal divisors $div(f) = sum_P ord_P(f) (P)$ form a subgroup of $Div(E)$; denote it by $PrDiv(E)$.
The divisor class group is the quotient $Cl(E)= Div(E)/PrDiv(E)$.
Obviosly we can define canonically a map $E(k) to Cl(E), p to (P)-(O)$ where $O$ is the special point (=neutral element).
How to show that this map determine the group law on $E(k)$.
algebraic-geometry elliptic-curves
algebraic-geometry elliptic-curves
asked Jan 11 at 21:24
KarlPeterKarlPeter
6081315
asked Jan 11 at 21:24
KarlPeterKarlPeter
6081315
edited Jan 11 at 21:32
KarlPeter
asked Jan 11 at 21:24
KarlPeterKarlPeter
6081315
asked Jan 11 at 21:24
KarlPeterKarlPeter
6081315
asked Jan 11 at 21:24
KarlPeterKarlPeter
6081315
6081315
2
$begingroup$
$(O)$ isn't a principal divisor. Principal divisors have degree zero.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 21:27
$begingroup$
@LordSharktheUnknown: ah sure sure, $(O)-(O)$ should be the prinipal one and is obviously one. Thank you. But the point is why in $Cl(E)$ the group law is already determined.
$endgroup$
– KarlPeter
Jan 11 at 21:31
$begingroup$
This is Proposition III.3.4 (p. 61) in Silverman. Have you looked there?
$endgroup$
– André 3000
Jan 11 at 21:33
$begingroup$
Every degree zero divisor is congruent, modulo principal divisors, to exactly one of the form $(P)-(O)$.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 21:33
$begingroup$
It suffices to show (or define) that $P+Q-R-O$ is a principal divisor iff $P + Q= R$ in the group law. From this then $sum_j P_j-Q_j = R-O+div(f)$ where $R = sum_j P_j-Q_j$ in the group law and $div(f) = O-R+sum_j P_j-Q_j$ where $f$ is meromorphic
$endgroup$
– reuns
Jan 11 at 22:05
|
show 7 more comments
2
$begingroup$
$(O)$ isn't a principal divisor. Principal divisors have degree zero.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 21:27
$begingroup$
@LordSharktheUnknown: ah sure sure, $(O)-(O)$ should be the prinipal one and is obviously one. Thank you. But the point is why in $Cl(E)$ the group law is already determined.
$endgroup$
– KarlPeter
Jan 11 at 21:31
$begingroup$
This is Proposition III.3.4 (p. 61) in Silverman. Have you looked there?
$endgroup$
– André 3000
Jan 11 at 21:33
$begingroup$
Every degree zero divisor is congruent, modulo principal divisors, to exactly one of the form $(P)-(O)$.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 21:33
$begingroup$
It suffices to show (or define) that $P+Q-R-O$ is a principal divisor iff $P + Q= R$ in the group law. From this then $sum_j P_j-Q_j = R-O+div(f)$ where $R = sum_j P_j-Q_j$ in the group law and $div(f) = O-R+sum_j P_j-Q_j$ where $f$ is meromorphic
$endgroup$
– reuns
Jan 11 at 22:05
2
2
$begingroup$
$(O)$ isn't a principal divisor. Principal divisors have degree zero.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 21:27
$begingroup$
$(O)$ isn't a principal divisor. Principal divisors have degree zero.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 21:27
$begingroup$
@LordSharktheUnknown: ah sure sure, $(O)-(O)$ should be the prinipal one and is obviously one. Thank you. But the point is why in $Cl(E)$ the group law is already determined.
$endgroup$
– KarlPeter
Jan 11 at 21:31
$begingroup$
@LordSharktheUnknown: ah sure sure, $(O)-(O)$ should be the prinipal one and is obviously one. Thank you. But the point is why in $Cl(E)$ the group law is already determined.
$endgroup$
– KarlPeter
Jan 11 at 21:31
$begingroup$
This is Proposition III.3.4 (p. 61) in Silverman. Have you looked there?
$endgroup$
– André 3000
Jan 11 at 21:33
$begingroup$
This is Proposition III.3.4 (p. 61) in Silverman. Have you looked there?
$endgroup$
– André 3000
Jan 11 at 21:33
$begingroup$
Every degree zero divisor is congruent, modulo principal divisors, to exactly one of the form $(P)-(O)$.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 21:33
$begingroup$
Every degree zero divisor is congruent, modulo principal divisors, to exactly one of the form $(P)-(O)$.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 21:33
$begingroup$
It suffices to show (or define) that $P+Q-R-O$ is a principal divisor iff $P + Q= R$ in the group law. From this then $sum_j P_j-Q_j = R-O+div(f)$ where $R = sum_j P_j-Q_j$ in the group law and $div(f) = O-R+sum_j P_j-Q_j$ where $f$ is meromorphic
$endgroup$
– reuns
Jan 11 at 22:05
$begingroup$
It suffices to show (or define) that $P+Q-R-O$ is a principal divisor iff $P + Q= R$ in the group law. From this then $sum_j P_j-Q_j = R-O+div(f)$ where $R = sum_j P_j-Q_j$ in the group law and $div(f) = O-R+sum_j P_j-Q_j$ where $f$ is meromorphic
$endgroup$
– reuns
Jan 11 at 22:05
|
show 7 more comments
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070361%2fgroup-law-elliptic-curve-with-divisor-class-group%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070361%2fgroup-law-elliptic-curve-with-divisor-class-group%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
$(O)$ isn't a principal divisor. Principal divisors have degree zero.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 21:27
$begingroup$
@LordSharktheUnknown: ah sure sure, $(O)-(O)$ should be the prinipal one and is obviously one. Thank you. But the point is why in $Cl(E)$ the group law is already determined.
$endgroup$
– KarlPeter
Jan 11 at 21:31
$begingroup$
This is Proposition III.3.4 (p. 61) in Silverman. Have you looked there?
$endgroup$
– André 3000
Jan 11 at 21:33
$begingroup$
Every degree zero divisor is congruent, modulo principal divisors, to exactly one of the form $(P)-(O)$.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 21:33
$begingroup$
It suffices to show (or define) that $P+Q-R-O$ is a principal divisor iff $P + Q= R$ in the group law. From this then $sum_j P_j-Q_j = R-O+div(f)$ where $R = sum_j P_j-Q_j$ in the group law and $div(f) = O-R+sum_j P_j-Q_j$ where $f$ is meromorphic
$endgroup$
– reuns
Jan 11 at 22:05